
Let \(D\)  be a division ring and \(R=D\bigoplus D\)  with componentwise multiplication. Clearly, \(R\)  is reduced, so \(R\)  is pointwise semicommutative. Define \(\sigma (h,w)=(w,h)\) . Then \(\sigma \)  is an automorphism of \(R\) . Let \(f(x)=(1,0)x\in R[x;\sigma ]\) . Observe that \(f(x)^2=0\)  but \(f(x)xf(x)\ne 0\) . Hence \(R[x;\sigma ]\)  is not pointwise commutative.

Take \(\mathbb {Z}_2\)  as the field of integers modulo 2 and let \(A=\mathbb {Z}_2[a_0,a_1,a_2,b_0, \\ b_1,b_2,c]\)  be the free algebra of polynomials with zero constant terms in noncommuting indeterminates \(a_0,a_1,a_2,b_0,b_1,b_2\)  and \(c\)  over \(\mathbb {Z}_2.\)  Take an ideal \(I\)  of the ring \(\mathbb {Z}_2+A\)  generated by \(a_0b_0, a_0b_1+a_1b_0, a_0b_2+a_1b_1+a_2b_0, a_1b_2+a_2b_1, a_1b_1,a_0rb_0, a_2rb_2, b_0a_0, b_0a_1+b_1a_0, b_0a_2+b_1a_1+b_2a_0, b_1a_2+b_2a_1,b_2a_2,b_0ra_0,b_2ra_2, (a_0+a_1+a_2)r(b_0+b_1+b_2),(b_0+b_1+b_2)r(a_0+a_1+a_2) ~and~ r_1r_2r_3r_4\)  where \(r,r_1,r_2,r_3,r_4\in A\) . Take \(R=(\mathbb {Z}_2+A)/I\) . Then we have \(R[x]\cong (\mathbb {Z}_2+A)[x]/I[x]\) . By [1]}, \(R\)  is reversible and hence pointwise semicommutative.
Observe that \((b_0+b_1x+b_2x^2)(a_0+a_1x+a_2x^2)\in I[x]\) . But \((b_0+b_1x+b_2x^2)c(a_0+a_1x+a_2x^2)\notin I[x]\) , since \(b_0ca_1+b_1ca_0\notin I\) . Hence \(l(\overline{(a_0+a_1x+a_2x^2)})\)  is not an ideal of \(R[x]\) . Again, \((a_0+a_1x+a_2x^2)(b_0+b_1x+b_2x^2)\in I[x]\) . But \((a_0+a_1x+a_2x^2)c(b_0+b_1x+b_2x^2)\notin I[x]\) , since \(a_0cb_1+a_1cb_0\notin I\) . Thus, \(r(\overline{(a_0+a_1x+a_2x^2)})\)  is also not an ideal of \(R[x]\) . Therefore, \(R[x]\)  is not pointwise semicommutative.


